int_0^3 (2+x^2) dx = | vedclass

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(C) ધારો કે $f(x) = 2+x^2$. નિશ્ચિત સંકલનની વ્યાખ્યા મુજબ, $\int_a^b f(x) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{bn} f\left(\frac{r}

Vedclass · Accepted Answer

$\lim _{n \rightarrow \infty} \frac{1}{n} \left[6n + \frac{1^2+2^2+\ldots+(3n)^2}{n^2} \right]$

Vedclass · Answer

(C) ધારો કે $f(x) = 2+x^2$. નિશ્ચિત સંકલનની વ્યાખ્યા મુજબ, $\int_a^b f(x) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{bn} f\left(\frac{r}{n}\right)$ અહીં, $\int_0^3 (2+x^2) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{3n} \left(2 + \left(\frac{r}{n}\right)^2\right)$ $= \lim_{n \rightarrow \infty} \frac{1}{n} \left[ \sum_{r=1}^{3n} 2 + \sum_{r=1}^{3n} \frac{r^2}{n^2} \right]$ $= \lim_{n \rightarrow \infty} \frac{1}{n} \left[ 2(3n) + \frac{1^2+2^2+\ldots+(3n)^2}{n^2} \right]$ $= \lim_{n \rightarrow \infty} \frac{1}{n} \left[ 6n + \frac{1^2+2^2+\ldots+(3n)^2}{n^2} \right]$.

$\int_0^3 (2+x^2) dx = $

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